Page 21 - RMGO 6

Rezolvarea problemelor din num˘arul anterior 21

Clasa a IX-a

MGO 181. Ar˘atat ,i c˘a pentru orice a, b, c ∈ R astfel ˆıncˆat ab + bc + ca = 3 avem

4a + 1 4b + 1 4c + 1
+ + ≤ 1.
2
2
2
7a + 4a + 4 7b + 4b + 4 7c + 4c + 4
Ardak Mirzakhmedov, Kazahstan
Solut ,ie. Inegalitatea din enunt , poate fi rescris˘a, succesiv:
8a + 2 8b + 2 8c + 2
+ + ≤ 2;
2
2
2
7a + 4a + 4 7b + 4b + 4 7c + 4c + 4
8a + 2 8b + 2 8c + 2
1 − + 1 − + 1 − ≥ 1;
2
2
2
7a + 4a + 4 7b + 4b + 4 7c + 4c + 4
2
2
2
5a + 2(a − 1) 2 5b + 2(b − 1) 2 5c + 2(c − 1) 2
+ + ≥ 1.
2
2
2
7a + 4a + 4 7b + 4b + 4 7c + 4c + 4
Este suficient s˘ demonstr˘am ultima inegualitate. Notˆand
a
2
2
2
2
2
2
5a + 2(a − 1) = x, 5b + 2(b − 1) = y, 5c + 2(c − 1) = z,
2
2
2
7a + 4a + 4 = m, 7b + 4b + 4 = n, 7c + 4c + 4 = k,
x y z
a
a
x, y, z, m, n, k > 0, trebuie s˘ ar˘at˘am c˘ + + ≥ 1. Dar, utilizˆand Inegalitatea
m √ n √ k √
x y z ( x + y + z) 2
Cauchy-Schwarz, avem + + ≥ , deci este suficient s˘a
n
m
√ √ √ 2 k m + n + k √ √ √
demonstr˘am c˘a x + y + z ≥ m + n + k, adic˘a 2 xy + 2 yz + 2 zx ≥
√ √ √
m + n + k − (x + y + z), adic˘a xy + yz + zx ≥ 4(a + b + c) + 3.
Utilizˆand Inegalitatea Cauchy-Schwarz, avem
√ √ »
2
2
2
2
xy ≥ 5a · 5b + 2(a − 1) · 2(b − 1) ≥ 5ab + 2(a − 1)(1 − b).
√ √
Analog, yz ≥ 5bc + 2(b − 1)(1 − c), zx ≥ 5ca + 2(c − 1)(1 − a), deci prin
√ √ √
adunare obt , inem xy + yz + zx ≥ 3(ab + bc + ca) + 4(a + b + c) − 6, deci
√ √ √
xy + yz + zx ≥ 4(a + b + c) + 3, ceea ce ˆıncheie demonstrat , ia.
MGO 182. Fie k > 1 s , i a, b, c, d > 0 astfel ˆıncˆat ab + bc + ca = d. Ar˘atat ,i c˘a
Ä p √ ä Ä p √ ä Ä p √ ä 3
2
2
2
a k − 1 + d b k − 1 + d c k − 1 + d ≤ k (a + b)(b + c)(c + a).
Cˆand are loc egalitatea?
Daniel Jinga, Pites , ti